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How to find the inverse of a function

  1. Apr 12, 2016 #1
    • moved into h/w help, so template is missing
    T(t) = Ts+(98.6 – Ts)e-kt

    rewrite in the form t=g-1(T)

    In trying to understand how to find the inverse of this but am having a hard time, please advise.

    Thanks,
    Kupkake303
     
    Last edited by a moderator: Apr 13, 2016
  2. jcsd
  3. Apr 12, 2016 #2

    stevendaryl

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    This is just a matter of trying to get [itex]t[/itex] by itself on one side of an equality. You initially have: [itex]T = T_s + (98.6 - T_s) e^{-kt}[/itex]. To simplify the expression, let [itex]U = e^{-kt}[/itex]. Then the equation is: [itex]T = T_s + (98.6 - T_s) U[/itex]. So solve for [itex]U[/itex]. Then solve for [itex]t[/itex] in terms of this value of [itex]U[/itex].
     
  4. Apr 13, 2016 #3

    So I would put it as 0=Ts+(98.6-Ts)U ??

    Doing that would give me -Ts/(98.6-Ts)=U

    Then putting the e-kt back in you get -Ts/(98.6-Ts)=e-kt

    So then I would then need to get that by it's self so the t is by its self but I can't remember how to do this. Please advise. Thank you
     
  5. Apr 13, 2016 #4

    stevendaryl

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    No, [itex]T = T_s + (98.6 - T_s)U[/itex]

    So [itex]U = (T-T_s)/(98.6 - T_s)[/itex]
     
  6. Apr 13, 2016 #5

    Right, because the T wouldn't just disappear.

    So next I would change the U back to a e-kt for e-kt=(T-Ts)/(98.6-Ts)

    from there would I try to figure out the e-kt ?
     
  7. Apr 13, 2016 #6

    stevendaryl

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    You want to take the natural log of both sides of the equation.
     
  8. Apr 13, 2016 #7
    okay so I would
    e-kt=(T-Ts)/(98.6-Ts)

    lne-kt=ln(T-Ts)/(98.6-Ts)

    The ln would cancel out the e

    -kt=ln(T-Ts)/(98.6-Ts)

    then divide by -k? to get

    t=(1/-k)ln(T-Ts)/(98.6-Ts)....?

    so this would be my answer for finding the inverse ..?
     
  9. Apr 13, 2016 #8

    stevendaryl

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    Yes, except that [itex]-ln(A/B) = ln(B/A)[/itex], so you can get rid of the [itex]-[/itex] sign to get:

    [itex]t = \frac{1}{k} ln((98.6 - T_s)/(T-T_s))[/itex]
     
  10. Apr 13, 2016 #9
    Awesome! Thank you soooo much. I spent about 4-5 hours trying to figure this problem out myself and I wouldn't have gotten through it by myself.
     
  11. Apr 13, 2016 #10

    epenguin

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    Logarithms are the inverse of the finish of exponentiation, by definition. Hopefully you remember your first lesson about them. Well hopefully they told you that then.
     
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